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\(a)\frac{7}{3}-\frac{4}{3}\times x=\frac{5}{6}\)
\(-\frac{4}{3}\times x=\frac{3}{2}\)
\(x=\frac{-9}{8}\)
\(b)\frac{7}{4}-\frac{3}{4}\times x=\frac{5}{6}\)
\(-\frac{3}{4}\times x=\frac{11}{12}\)
\(x=-\frac{11}{9}\)
2/3.x + 1/4 = 7/12
2/3.x = 7/12 - 1/4
2/3.x = 1/3
x = 1/3 : 2/3
x = 1/2
Bài làm
\(\frac{2}{3}x+\frac{1}{4}=\frac{7}{12}\)
\(\frac{2}{3}x=\frac{7}{12}-\frac{1}{4}\)
\(\frac{2}{3}x=\frac{7}{12}-\frac{3}{12}\)
\(\frac{2}{3}x=\frac{4}{12}\)
\(\frac{2}{3}x=\frac{1}{3}\)
\(x=\frac{1}{3}:\frac{2}{3}\)
\(x=\frac{1}{3}.\frac{3}{2}\)
\(x=\frac{1}{2}\)
Vậy \(x=\frac{1}{2}\)
a) (4 - x)3 = -125
(4 - x)3 = (-5)3
\(\Rightarrow4-x=-5\)
\(\Rightarrow x=4-\left(-5\right)\)
\(\Rightarrow x=9\)
Vậy x = 9
b) (x - 5).(x + 7) = 0
\(\Rightarrow\hept{\begin{cases}x-5=0\\x=0+5\\x=5\end{cases}hoac\hept{\begin{cases}x+7=0\\x=0-7\\x=-7\end{cases}}}\)
Vậy\(x\in\left\{5;-7\right\}\)
Chúc bạn học tốt !
\(\left(4-x\right)^3\)\(=\)\(-125\)
\(\Leftrightarrow\)\(\left(4-x\right)^3\)\(=\)\(\left(-5\right)^3\)
\(\Leftrightarrow\)\(4-x=-5\)
\(\Leftrightarrow\)\(x=4-\left(-5\right)\)
\(\Leftrightarrow\)\(x=4+5=9\)
Vậy \(x=9\)
\((x-5)\left(x+7\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x-5=0\\x+7=0\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=5\\x=-7\end{cases}}\)
Vậy \(x\in\text{ }\left\{5;-7\right\}\)
`Answer:`
\(\left(\frac{3}{10}-x\right)-\frac{1}{3}=-\frac{2}{5}\)
\(\Leftrightarrow\frac{3}{10}-x=-\frac{2}{5}:-\frac{1}{3}\)
\(\Leftrightarrow\frac{3}{10}-x=\frac{6}{5}\)
\(\Leftrightarrow x=\frac{3}{10}-\frac{6}{5}\)
\(\Leftrightarrow x=-\frac{9}{10}\)
\(\frac{7}{9}-\frac{2}{9}x=\frac{4}{3}\)
\(\Leftrightarrow\frac{2}{9}x=\frac{7}{9}-\frac{4}{3}\)
\(\Leftrightarrow\frac{2}{9}x=-\frac{5}{9}\)
\(\Leftrightarrow x=-\frac{5}{9}:\frac{2}{9}\)
\(\Leftrightarrow x=-\frac{5}{2}\)
\(\frac{1}{3}+\frac{2}{3}:x=-7\)
\(\Leftrightarrow\frac{2}{3}:x=-7-\frac{1}{3}\)
\(\Leftrightarrow\frac{2}{3}:x=-\frac{22}{3}\)
\(\Leftrightarrow x=-\frac{22}{3}.\frac{2}{3}\)
\(\Leftrightarrow x=-\frac{44}{9}\)
b) \(\left(x+1\right)+\left(x+4\right)+...+\left(x+28\right)=155\)
\(\Leftrightarrow10x+\left(1+4+...+28\right)=155\)
\(\Leftrightarrow10x+145=155\)
\(\Leftrightarrow10x=10\)
\(\Leftrightarrow x=1\)
Vậy x=1
a) \(\frac{3x}{-4}=\frac{-9}{6}\)
=> \(\frac{-3x}{4}=\frac{-9}{6}\)
=> \(\frac{-3x}{4}=\frac{-3}{2}\)
=> \(\frac{-3x}{4}=\frac{-6}{4}\)
=> \(-3x=-6\)
=> \(x=\left(-6\right):\left(-3\right)=2\)
b) \(\frac{2}{x}=\frac{y}{10}=\frac{4}{-5}\)
=> \(\frac{2}{x}=\frac{y}{10}=\frac{-4}{5}\)
+) \(\frac{2}{x}=\frac{-4}{5}\)
=> \(\left(-4\right)\cdot x=2\cdot5\)
=> \(\left(-4\right)\cdot x=10\)
=> \(x=\frac{10}{-4}=\frac{-10}{4}=\frac{-5}{2}\)
+) \(\frac{y}{10}=\frac{-4}{5}\)
=> \(5\cdot y=-40\)
=> \(y=\left(-40\right):5=-8\)
Vậy \(x=-\frac{5}{2},y=-8\)
c) \(\frac{7}{-x}=\frac{5}{y}=\frac{2}{34}\)
=> \(\frac{-7}{x}=\frac{5}{y}=\frac{1}{17}\)
+) \(\frac{-7}{x}=\frac{1}{17}\Rightarrow x=\left(-7\right)\cdot17=-119\)
+) \(\frac{5}{y}=\frac{1}{17}\Rightarrow y=5\cdot17=85\)
Vậy \(x=-119,y=85\)
d) \(\frac{x^2}{4}=16\)
=> \(x^2=16\cdot4=64\)
=> \(x\in\left\{8;-8\right\}\)
e) \(\frac{3}{x^3}=\frac{4}{-36}\)
=> \(\frac{3}{x^3}=\frac{1}{-9}=\frac{-1}{9}\)
=> \(x^3=3:\left(-\frac{1}{9}\right)=-27\)
=> \(x=-3\)
Vậy x = -3
\(b,8-|x-7|=2\)
\(|x-7|=6\)
\(\Rightarrow\orbr{\begin{cases}x-7=6\\x-7=-6\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=13\\x=1\end{cases}}\)
\(\frac{x-7}{3}=\frac{x}{-4}=\frac{y}{7}=\frac{4}{-t2}\)
\(\Leftrightarrow\frac{x-7}{3}=\frac{x}{-4}=\frac{y}{7}=\frac{2}{t}\)
3 = 3
4 = 2^2
7 = 7
MC = 3. 2^2 . 7 = 84
\(\Leftrightarrow\frac{28\left(x-7\right)}{84}=\frac{-21x}{84}=\frac{12y}{84}=\frac{2}{t}\)
\(\Leftrightarrow28x-196=-21x=12y=\frac{2}{t}\)
\(\Leftrightarrow\frac{28xt-196t}{t}=\frac{-21xt}{t}=\frac{12yt}{t}=\frac{2}{t}\)
\(\Leftrightarrow28xt-196t=-21xt=12yt=2\)